AnExtremelyLongName

Yay I'm back after months of video games and coding.

What DOESN'T appear one time? We can call that being "deleted".
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Let's call a term in list A "\(n\)"

 - \(n\) is only deleted if \(n\text{ }mod\text{ }4=3 \) and not \(n + 1\text{ }mod\text{ }8=7\)

so in other words, if \(n\) is a multiple of 3 and \(n+1\) is NOT a multiple of 7, then it shall be deleted.

Example {n = 3}

 - n is multiple of 3

 - n + 1 is NOT multiple of 7

therefore it should be deleted.

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Now let us calculate how much will be deleted.

There are floor(1000/3) = 333 multiples of 3 from 1 - 1000.

There are floor(333/7) = 47 multiples of 7 that are also a multiple of 3.

That means there are 333 - 47 = 286 multiples of 3 that are NOT multiples of 7. That means 286 should be deleted.

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Since the question asks how many numbers are not deleted in list C, we can easily find that the answer is 1000 - 286 is \(\boxed{714}\).

Please correct me if I am wrong.

Group every two terms and factor (a,b,c) respectively:

\(a(b^3-a^2b)+b(c^3-b^2c)+c(a^3-c^2a)\)

Now factor (b,c,a) respectively:

\(ab(b^2-a^2)+bc(c^2-b^2)+ca(a^2-c^2)\)

Now obviously follow up to factor the subtraction of two perfect squares.

\(ab(b-a)(b+a)+bc(c-b)(c+b)+ca(a-c)(a+c)\)

Not sure if this can be factored further

Your latex sucks.

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 - Just plug into the distance formula and solve.

\((2\sqrt{10})^2=40\)

\((4-2a)^2+(-1-(a-4))^2=40\)

\((4-2a)^2+(3-a)^2=40\)

Now solve (if you can, try using vieta's formulas to do it even faster.)

I'm bored so why not

\((r-8)^2+(r-9)^2=r^2\)

Solve for \(r\)

Yes, I know this is a problem. I too, am a user on AoPS and I have seen so much cheating on this site. (I am not enrolled in classes though). The problem is, we can't really recongize which problems are from AoPS so we kind of need help from users like you. 

One question per post please.

1. This is mostly geometry believe it or not.

The value of K in the circle equation would be the blue line squared. Or in other words, the blue line is \(\sqrt{k}\)

The value of BC in the tangent equation would just be \(k\)

We know ABC is a 45 - 45 - 90 triangle (If the coefficients of x and y are 1 when graphed in standard form for the red line). The blue line is the perpendicular bisector. 

Based on pythagorean theorem: \(2 *\sqrt{k}^2 = k^2\) (two legs are \(\sqrt{k}\) and hypotenuse is BC)

\(2k=k^2\)

\(k(k-2)=0\)

\(\boxed{k=2}\)

1. Connect \(RG\)

2. Since \(FR=FG\), triangle \(FRG\) is isosceles with \(F\) as the vertex.

3. Angle \(GRF\) subtends arc \(GF\), which is 130, therefore angle \(GRF\) = 130/2 = \(65\)

4. Since \(FRG\) is isosceles, that means \(FRG=RGF\).

5. Angle \(GFR\) = \(50\) because all triangles have an angle sum of 180, and we know that two angles are 65 degrees. (180 - 65 * 2)

6. Angle \(PGF\) subtends arc \(FQ\), so angle \(PGF\) = 28/2 = \(14\)

7. Conside triangle \(PFG\), we know the measure of two angles \(GFR\) and \(PGF\) is  \(50\) and \(14\), respectively. That means the third angle \(GPF\) is 180 - 50 - 14 = \(116\)

8. By supplements angle \(RPG=180-116 = 64\)

Done

Just remember \(f^{-1}\) is just swapping \(x\) and \(y\). 

An alternative method is draw the altitude from C to AB. Then calculate the area, and I got 31.25 as well.